Question:

If $(x)^y = (y)^x$, then find $\frac{dy}{dx}$.

Show Hint

When differentiating implicit functions involving logarithms, always apply the chain rule and product rule carefully to each term.
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

We are given the equation: \[ x^y = y^x. \] Taking the natural logarithm of both sides: \[ \log(x^y) = \log(y^x). \] Using the logarithmic identity $\log(a^b) = b \log(a)$, we get: \[ y \log x = x \log y. \] Now, differentiate both sides with respect to $x$. On the left-hand side, apply the product rule to $y \log x$, and on the right-hand side, apply the product rule to $x \log y$: \[ \frac{d}{dx}(y \log x) = \frac{d}{dx}(x \log y). \] Differentiating: \[ \frac{dy}{dx} \log x + y \frac{1}{x} = \frac{dy}{dx} x \frac{1}{y} + \log y. \] Now, collect terms involving $\frac{dy}{dx}$ on one side: \[ \frac{dy}{dx} \log x - \frac{dy}{dx} \frac{x}{y} = \log y - \frac{y}{x}. \] Factor out $\frac{dy}{dx}$: \[ \frac{dy}{dx} \left( \log x - \frac{x}{y} \right) = \log y - \frac{y}{x}. \] Finally, solve for $\frac{dy}{dx}$: \[ \frac{dy}{dx} = \frac{\log y - \frac{y}{x}}{\log x - \frac{x}{y}}. \]
Was this answer helpful?
1
0